HDU 3415 Max Sum of Max-K-sub-sequence（单调队列）

Given a circle sequence A11,A22,A33......Ann. Circle sequence means the left neighbour of A11 is Ann , and the right neighbour of Ann is A11. 
Now your job is to calculate the max sum of a Max-K-sub-sequence. Max-K-sub-sequence means a continuous non-empty sub-sequence which length not exceed K.

Input

The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. 
Then T lines follow, each line starts with two integers N , K(1<=N<=100000 , 1<=K<=N), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output a line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the minimum start position, if still more than one , output the minimum length of them.

Sample Input

4
6 3
6 -1 2 -6 5 -5
6 4
6 -1 2 -6 5 -5
6 3
-1 2 -6 5 -5 6
6 6
-1 -1 -1 -1 -1 -1

   【题意】一个长度为n的循环序列，在其中找一个长度为k(k<=n)的子序列，要求这个子序列的和尽可能大，最后输出最大值，并且输出子序列的起点和终点。

   【分析】先用sum数组来保存前i个数的和，记得还要保存i到n+k的数的和，因为这是要构成首尾相连。然后从头开始，找子序列的最大值，双端单调队列队首永远是序列和最大的，往后依次递减，每移动一次保存一下和前一次ans的最小值，最后输出。

   【用时】452 ms！！

   【AC代码】

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N=2e5+10;
const int INF=1e7;

int a[N],sum[N];
int head;
int tail;

int str[N];
int n,nn,k,maxn;
int s,e;

int min(int x,int y)
{
    return x>y?y:x;
}

void push_up(int i)
{
    str[tail++]=i;
}

bool isempty()
{
    return head==tail;
}

int Front()
{
    return str[head];
}

void Pop_back()
{
    tail--;
}

void Pop_front()
{
    head++;
}

void solve()
{
    head=tail=0;
    int i;
    for(i=1;i<=n;i++)
    {
        while(!isempty() && sum[i-1]<sum[str[tail-1]]) Pop_back();
        while(!isempty() && str[head]+k<i) Pop_front();
        push_up(i-1);
        if(sum[i]-sum[str[head]]>maxn)
        {
            maxn = sum[i]-sum[str[head]];
            s = str[head]+1;
            e= i ;
        }
    }
    if(e>nn)
        e%=nn;
    printf("%d %d %d\n",maxn,s,e);
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
       scanf("%d%d",&n,&k);
       nn=n;
       int i;
       for(i=1,sum[0]=0;i<=n;i++)
       {
           scanf("%d",&a[i]);
           sum[i]=sum[i-1]+a[i];
       }
       for(;i<n+k;i++)
           sum[i]=sum[i-1]+a[i-n];
       n=n+k-1;
       maxn=-INF;
       solve();
    }
    return 0;
}